Welcome to March 2017

Algebra Level 1

Playing with the numbers in today's date (March 2017), I started to write out this equation:

3 20 − 3 17 = − 3 20 × 3 17 \frac{ {\color{#3D99F6}3}} { {\color{#20A900}20}} - \frac{ {\color{#3D99F6}3}}{ {\color{#D61F06}17}} = - \frac{{\color{#3D99F6}3}}{{\color{#20A900}20}} \times \frac{{\color{#3D99F6}3}}{{\color{#D61F06}17}}

Is the equation true or false?

True False

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Chung Kevin by Chung Kevin , 45, Australia

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4 solutions

Jesse Nieminen
Mar 5, 2017

3 20 − 3 17 = 3 ⋅ 17 20 ⋅ 17 − 3 ⋅ 20 17 ⋅ 20 = 3 ⋅ 17 − 3 ⋅ 20 20 ⋅ 17 = 3 ( 17 − 20 ) 20 ⋅ 17 = − 3 ⋅ 3 20 ⋅ 17 = − 3 20 × 3 17 \dfrac{\color{#3D99F6}3}{\color{#20A900}20}-\dfrac{\color{#3D99F6}3}{\color{#D61F06}17}=\dfrac{\color{#3D99F6}3\color{#333333}\cdot\color{#D61F06}17}{\color{#20A900}20\color{#333333}\cdot\color{#D61F06}17}-\dfrac{\color{#3D99F6}3\color{#333333}\cdot\color{#20A900}20}{\color{#D61F06}17\color{#333333}\cdot\color{#20A900}20}=\dfrac{\color{#3D99F6}3\color{#333333}\cdot\color{#D61F06}17\color{#333333}-\color{#3D99F6}3\color{#333333}\cdot\color{#20A900}20}{\color{#20A900}20\color{#333333}\cdot\color{#D61F06}17}=\dfrac{\color{#3D99F6}3\color{#333333}\left(\color{#D61F06}17\color{#333333}-\color{#20A900}20\right)}{\color{#20A900}20\color{#333333}\cdot\color{#D61F06}17}=-\dfrac{\color{#3D99F6}3\color{#333333}\cdot\color{#3D99F6}3}{\color{#20A900}20\color{#333333}\cdot\color{#D61F06}17}=-\dfrac{\color{#3D99F6}3}{\color{#20A900}20}\times\dfrac{\color{#3D99F6}3}{\color{#D61F06}17}

Hence, the answer is True \boxed{\text{True}} .

58 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

The comments to this solution mention the general problem: For what values of a a and b b is it true that a + b = a b ? a + b = ab ?

-0.02647058823

Mark Mai Woerds - 4 years, 3 months ago

Nicely done! It is fascinating to see that the sum and the product of 3 20 \frac{3}{20} and − 3 17 - \frac{3}{17} are equal, since it is not generally true that the sum of two numbers is equal to their product.

How could we go about finding all pairs of numbers a , b a, b , whose sum is equal to their product a + b = a b a + b = ab ?

Pranshu Gaba - 4 years, 3 months ago

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a + b = a b a b − a − b = 0 a b − a − b + 1 = 1 ( a − 1 ) ( b − 1 ) = 1 \begin{aligned} a + b &= ab \\ ab - a - b &= 0 \\ ab - a - b + 1 &= 1 \\ \left(a-1\right)\left(b-1\right) &= 1 \\ \end{aligned}

Now if a a and b b are integers, a − 1 a-1 and b − 1 b-1 cannot have any prime factors and thus are either 1 1 or − 1 -1 .

Thus, the only ordered pairs ( a , b ) \left(a,b\right) which satisfy the condition are ( 0 , 0 ) \left(0,0\right) and ( 2 , 2 ) \left(2,2\right) .

Or we could see that ( 0 , 0 ) \left(0,0\right) is trivial and then divide both sides by a b ab and then have 1 a + 1 b = 1 \dfrac1a + \dfrac1b = 1 .
Which forces a < 3 ∨ b < 3 a < 3 \vee b < 3 which again forces a = b = 2 a = b = 2 .

Also, all rational solutions can be found by picking arbitrary rational a a in and solving for b b .

Same can be done for real and complex (, quoternion, ...) solutions.

Jesse Nieminen - 4 years, 3 months ago

We can say 1 + 2 + 3 = 1 × 2 × 3 1+2+3=1 \times 2 \times 3 this shows a + b + c = a b c a+b+c=abc but this does not applies for all

Jehova Jesus Thomas - 4 years, 3 months ago

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yeah, a worthwhile exercise is to explain why this equation is not always true.

Agnishom Chattopadhyay - 4 years, 3 months ago

a = b / (b-1) for all possible solution

lu picc - 3 years, 3 months ago

More generally, under what conditions is it true that a b − c d = a b × c d ? \frac ab - \frac cd = \frac ab \times \frac cd? Algebraic manipulation shows that this is true iff ( a + b ) ( d − c ) = b d . (a + b)(d - c) = bd. In this case, a = c = 3 a = c = 3 , b = 17 b = 17 and d = 20 d = 20 , and obviously 20 ⋅ 17 = 17 ⋅ 20 20\cdot 17 = 17\cdot 20 . We will have the same situation in February 2018 or January 2019...

(Note: I rewrote the equation without negative sign on the right, by flipping the two fractions in the subtraction.)

40 Helpful 0 Interesting 0 Brilliant 0 Confused

Great job on generalizing. Is it true that there are infinitely many integral solutions to this equation?

Agnishom Chattopadhyay - 4 years, 3 months ago

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One infinite family of solutions is generated by a = c = n a = c = n , b = m b = m , d = m + n d = m + n for n , m ∈ N n,m \in \mathbb N . I.e. n m − n m + n = n m × n m + n . \frac nm - \frac n{m+n} = \frac nm \times \frac n{m+n}. To get irreducible fractions we need m m and n n , as well as n n and n + m n+m , to be coprime. There are still infinitely many solutions.

Arjen Vreugdenhil - 4 years, 3 months ago
Zach Abueg
Mar 1, 2017

3 20 − 3 17 = 3 ( 17 ) − 3 ( 20 ) 20 × 17 = 51 − 60 20 × 17 = − 9 20 × 17 \displaystyle \frac {3}{20} - \frac {3}{17} = \frac {3(17) - 3(20)}{20 \times 17} = \frac {51 - 60}{20 \times 17} = \frac {-9}{20 \times 17}

− 9 20 × 17 = − 3 × 3 20 × 17 \displaystyle \frac {-9}{20 \times 17} = \frac {-3 \times 3}{20 \times 17}

− 9 = − 3 × 3 \displaystyle -9 = -3 \times 3

− 9 = − 9 \displaystyle -9 = -9

✔ \displaystyle ✔

29 Helpful 0 Interesting 0 Brilliant 0 Confused

In general, subtracting (or adding) fractions is not the same as multiplying them. However, there can be unique scenarios where this comes through.

What a nice result to share about 03/2017!

Calvin Lin Staff - 4 years, 3 months ago

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Waiting for 02/2018, 01/2019, 01/2021, 02/2022, ... :D

Jesse Nieminen - 4 years, 3 months ago
Viki Zeta
Mar 5, 2017

3 20 − 3 17 = 3 × ( 1 20 − 1 17 ) = 3 ( 17 − 20 17 × 20 ) = 3 × − 3 17 × 20 = − 3 20 × 3 17 \dfrac{3}{20} - \dfrac{3}{17} \\ = 3 \times \left(\dfrac{1}{20} - \dfrac{1}{17}\right) \\ = 3 \left( \dfrac{17 - 20}{17 \times 20}\right) \\ = 3 \times \dfrac{-3}{17 \times 20} \\ = - \dfrac{3}{20} \times \dfrac{3}{17}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Good and clear sequence of steps!

Beautiful equation, don't you think?

Pi Han Goh - 4 years, 3 months ago

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Ya I think.... Solving it is easy but with this a problem can be made in general

Md Zuhair - 4 years, 3 months ago

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